# Alexandrov's rigidity in higher dimensions

If $$\Phi_1,\Phi_2$$ are convex polyhedra in $$\mathbb{R}^3$$ such that the sets of outer normals to facets coincide, but $$\Phi_1$$ is not a translate of $$\Phi_2$$, then there exist two corresponding facets $$F_1,F_2$$ (with the same outer normal) such that one of them is a translate of a proper subset of another.

This is A. D. Alexandrov's theorem, which generalizes the theorem of H. Minkowski which assumes that the areas of corresponding facets are always equal.

If I remember well, then in dimensions greater than 3 this is no longer true (while Minkowski theorem holds true in any dimension.) The request is a reference to counterexamples.

Here is a counterexample in dimension four.

Consider positive numbers $$x_1,x_2,x_3,x_4\in\Bbb R$$ with $$x_1 and $$x_3 and construct the two 4-orthotopes (cartesian products of intervals)

\begin{align} O_1:=[0,x_1]\times [0,x_2]\times[0,x_3]\times[0,x_4]\\ O_2:=[0,x_2]\times [0,x_1]\times[0,x_4]\times[0,x_3] \end{align}

A pair of parallel facets is defined by a 3-element subset $$I=\{i_1,i_2,i_3\}\subset \{1,2,3,4\}$$:

\begin{align} F_1&:=[0,x_{i_1}]\times[0,x_{i_2}]\times[0,x_{i_3}] \subset O_1 \\ F_2&:=[0,x_{\sigma(i_1)}]\times[0,x_{\sigma(i_2)}]\times[0,x_{\sigma(i_3)}] \subset O_2 \end{align}

where $$\sigma$$ is the permutation $$(12)(34)$$. (strictly spoken, the inclusions are wrong, but I hope the idea is clear).

For $$F_1$$ to be (parallel to) a subset of $$F_2$$ it must hold $$(*)\,x_i\le x_{\sigma(i)}$$ for all $$i\in I$$. But each 3-element subset $$I\subset\{1,2,3,4\}$$ contains either $$\{1,2\}$$ or $$\{3,4\}$$, and so $$(*)$$ cannot be satisfied.

The easiest example is probably $$(x_1,x_2,x_3,x_4)=(1,2,1,2)$$.

• Great, and I had a look at Alexandrov's book and found the same example there. – Fedor Petrov Jan 14 at 17:50